The derivative of \(y=\tan ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin…
The derivative of \(y=\tan ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] \text { with }\) respect to \(x\) is equal to
-1
0
\pm 2
\(\pm \frac{1}{2}\)
Solution
It is given that,
\(\begin{aligned}
& y=\tan ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] \\
& \because \quad \sqrt{1+\sin x}=\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\left|\cos \frac{x}{2}+\sin x / 2\right|
\end{aligned}\)
and \(\sqrt{1-\sin x}=\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\)
\(=\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|\)
\(\therefore \quad y=\tan ^{-1}\left(\cot \frac{x}{2}\right)\)
when \(\cos \frac{x}{2}+\sin \frac{x}{2}\)
and \(\cos \frac{x}{2}-\sin \frac{x}{2}=\frac{\pi}{2}-\frac{x}{2}\) are positive
\(\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\),
when \(\cos \frac{x}{2}+\sin \frac{x}{2}\) is positive and \(\cos \frac{x}{2}-\sin \frac{x}{2}\) is negative
\(\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\),
when \(\cos \frac{x}{2}+\sin \frac{x}{2}\) is negative
and \(\cos \frac{x}{2}-\sin \frac{x}{2}\) is positive
\(\tan ^{-1}\left(\cot \frac{x}{2}\right)=\frac{\pi}{2}-\frac{x}{2}\),
when \(\cos \frac{x}{2}+\sin \frac{x}{2}\) and
as \(\cos \frac{x}{2}-\sin \frac{x}{2}\) is negative.
So, \(\frac{d y}{d x}= \pm \frac{1}{2}\)
Hence, option (d) is correct.